Answer: 3400
Step-by-step explanation:
Given
Magnetic field, B = 0.1 T
Diameter of magnet, d = 2 cm = 0.02 m
Length of magnet, l = 8 cm = 0.08 m
Current of the magnet, I = 1.9 A
Number of turns needed, N = ?
To solve this problem, we would use the formula,
N = (LB) / (μI), where
μ = 1.257*10^-6 Tm/A, so that
N = (0.08 * 0.1) / (1.257*10^-6 * 1.9)
N = 0.008 / 2.388*10^-6
N = 3350
N ~ 3400
Therefore, the number of turns of wire needed is 3400